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1 year ago

1. Related to the number of particles in a gram of

Chemistry

1 answer:

Romashka-Z-Leto [24]1 year ago

6 0

Answer:

Avogadro's number or Avogardro’s constant

Explanation:

I’m pretty sure this is correct if it’s not I’m sorry lol.

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Why is there a delay for the tides to react to the Moon's and Sun's gravitational effect?

Sloan [31]

There is no delay because the effects of gravity are constantly acting on bodies of water. Hence, option D is correct.

<h3>What is the gravitational effect?</h3>

Gravitation or just gravity is the force of attraction between any two bodies.

The Moon exerts over twice the gravitational pull of the Sun on Earth's tides because of its proximity to Earth. -The lunar tidal bulges are about twice the size of the solar tidal bulges. -The Sun's contribution to the tides is less than that of the Moon.

When the sun and moon are perfectly unaligned (they form a 90-degree angle relative to the earth), there are still tides because the moon's gravitational gradient is stronger than the sun's. The sun's gravitational gradient never completely cancels out the moon's.

Hence, option D is correct.

Learn more about the gravitational effect here:

brainly.com/question/14433868

#SPJ1

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1 year ago

An object at a distance of 30 cm from a concave mirror gets its image at the same point. The focal length of the mirror is

Nimfa-mama [501]

The focal length of the mirror is 15 cm

From the question given above, the following data were obtained:

Object distance (u) = 30 cm

Image distance (v) = 30 cm

<h3>Focal length (f) =? </h3>

The focal length of the mirror can be obtained as follow:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\\\\frac{1}{f} = \frac{1}{30} + \frac{1}{30}\\\\\frac{1}{f} = \frac{2}{30} \\\\\frac{1}{f} = \frac{1}{15} \\\\$

<h3>Invert </h3><h3>f = 15 cm</h3>

Therefore, the focal length of the mirror is 15 cm

Learn more: brainly.com/question/1392083

3 0

2 years ago

Which of the following is NOT a sign of a chemical reaction:

Sergeeva-Olga [200]

The answer is B. Precipitation

7 0

3 years ago

Read 2 more answers

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction = a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0

2 years ago

How many grams of NaOH are<br> needed to make 400. mL of<br> 1.50 M solution?

juin [17]

Answer:

24g of NaOH are required

Explanation:

Molarity, M, is an unit of concentration widely used in chemistry defined as the ratio between moles of solute (In this case, NaOH), and volume of solution in liters.

We can find the moles of NaOH and its mass with the volume and desired concentration as follows:

<em>Moles NaOH:</em>

400.0mL = 0.400L * (1.50mol / L) = 0.600 moles NaOH

<em>Mass NaOH -Molar mass: 40.0g/mol-:</em>

0.600 moles * (40.0g / mol) =

<h3>24g of NaOH are required</h3>

7 0

2 years ago